Difference between revisions of "2021 AMC 10A Problems/Problem 21"
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Let <math>ABCDEF</math> be an equiangular hexagon. The lines <math>AB, CD,</math> and <math>EF</math> determine a triangle with area <math>192\sqrt{3}</math>, and the lines <math>BC, DE,</math> and <math>FA</math> determine a triangle with area <math>324\sqrt{3}</math>. The perimeter of hexagon <math>ABCDEF</math> can be expressed as <math>m +n\sqrt{p}</math>, where <math>m, n,</math> and <math>p</math> are positive integers and <math>p</math> is not divisible by the square of any prime. What is <math>m + n + p</math>? | Let <math>ABCDEF</math> be an equiangular hexagon. The lines <math>AB, CD,</math> and <math>EF</math> determine a triangle with area <math>192\sqrt{3}</math>, and the lines <math>BC, DE,</math> and <math>FA</math> determine a triangle with area <math>324\sqrt{3}</math>. The perimeter of hexagon <math>ABCDEF</math> can be expressed as <math>m +n\sqrt{p}</math>, where <math>m, n,</math> and <math>p</math> are positive integers and <math>p</math> is not divisible by the square of any prime. What is <math>m + n + p</math>? | ||
Revision as of 11:50, 12 February 2021
Contents
Problem
Let 
be an equiangular hexagon. The lines 
and 
determine a triangle with area 
, and the lines 
and 
determine a triangle with area 
. The perimeter of hexagon can be expressed as 
, where 
and 
are positive integers and is not divisible by the square of any prime. What is 
?
Solution (Misplaced problem?)
Note that the extensions of the given lines will determine an equilateral triangle because the hexagon is equiangular. The area of the first triangle is , so the side length is 
. The area of the second triangle is , so the side length is 
. We can set the first value equal to 
and the second equal to 
by substituting some lengths in with different sides of the same equilateral triangle. The perimeter of the hexagon is just the sum of these two, which is 
and 
Video Solution by OmegaLearn (Angle Chasing and Equilateral Triangles)
~ pi_is_3.14
See Also
| 2021 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
